A radioactive source emits 4 particles on average during a five-second period. Calculate the probability...

Question

A radioactive source emits 4 particles on average during a five-second period.

a) Calculate the probability that it emits 3 particles during a 5-second period.

$P(x = 3$) = $\frac{e^{-4} \times 4^3}{3!} = 0.1945$

b) Calculate the probability that it emits at least one particle during a 5- second period.

$P(x \geq 1$) = $\frac{e^{-4} \times 4^1}{1!} + \frac{e^{-4} \times 4^2}{2!} + \frac{e^{-4} \times 4^3}{3!} + \frac{e^{-4} \times 4^4}{4!} = 0.62$

I thought that I could do $P(x \geq 1$) = $1 - P(x \lt 1$) = $\frac{e^{-4} \times 4^0}{0!}$, but I get an answer of $0.98$. I'm not sure where I went wrong.

c) During a ten-second period, what is the probability that 6 particles are emitted?

Since the time period has doubled, I thought I could also double the average.

$P(x = 6$) = $ \frac{e^{-8} \times 8^6}{6!} = 0.12 $

Answer 1

For part $b)$

You need $P(X\ge1)=1-P(X=0)=1-\dfrac{e^{-4}\times4^0}{0!}=0.9817$

And your part $c)$ is correct.

Answer 2

If you just do P(X>=1) where x = 1,2,3,4 you wont get the rest of the probability where x>4, poisson distribution has no upper or lower limit so it should be done through the method of 1-P(X=0) instead to exclude P(X=0), getting P(X>=1).